Question

A small-signal voltage $V\left(t\right)=V_o\sin \omega t$ is applied across an ideal capacitor C

• A. Current I (t) is in phase with voltage V(t).
• B. Over a full cycle the capacitor C does not consume any energy from the voltage source.
• C. Current I(t), lags voltage V(t) by 90 degrees.
• D. Current I(t) leads voltage V(t) by 180 degrees.

Answer 1

The correct option is B

Over a full cycle the capacitor C does not consume any energy from the voltage source.

Given data

1. The small-signal voltage is $V\left(t\right)=V_o\sin \omega t$.

Explanation for the correct option

Option (B)

1. We know the power in an ac circuit is $P=I_{rms}.V_{rms}.\cos \theta$, where, $I_{rms}$ and $V_{rms}$ are the RMS value of current and voltage $\left(I_{rms}=\frac{I}{\sqrt{2}}and{V}_{rms}=\frac{V}{\sqrt{2}}\right).$
2. In a pure capacitor circuit, the current leads voltage by 90 degrees. So, $P=I_{rms}.V_{rms}.\cos \theta =0(\sin ce,\cos 90=0)$.
3. A purely capacitive circuit is a wattles circuit and over a full cycle, the capacitor C does not consume any energy from the voltage source.

Explanation for the incorrect options

1. The phase difference of current and voltage is 90 degrees in a purely capacitive circuit. So, option A is not correct.
2. 
3. So, options (A), (B), and (C) are incorrect.

Therefore, option (B) is correct.